Common Core: 6th Grade Math : Fluently Divide Multi-Digit Numbers: CCSS.Math.Content.6.NS.B.2

Study concepts, example questions & explanations for Common Core: 6th Grade Math

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Example Questions

Example Question #41 : The Number System

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 31\\ 5{\overline{\smash{)}155}}\\ -\ 15 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 15\)

\(\displaystyle 5\times 41\)

\(\displaystyle 5\times 21\)

\(\displaystyle 5\times 51\)

\(\displaystyle 5\times 31\)

Correct answer:

\(\displaystyle 5\times 31\)

Explanation:

The computation shows that \(\displaystyle 155\div5=31\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 5}{\overline{\smash{)}155}}\\ -\ 15 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times31=155+0\)

Simplify.

\(\displaystyle 5\times31=155\)

The correct answer is \(\displaystyle 5\times 31\)

Example Question #42 : The Number System

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 41\\ 5{\overline{\smash{)}205}}\\ -\ 20 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 31\)

\(\displaystyle 5\times 61\)

\(\displaystyle 5\times 62\)

\(\displaystyle 5\times 20\)

\(\displaystyle 5\times 41\)

Correct answer:

\(\displaystyle 5\times 41\)

Explanation:

The computation shows that \(\displaystyle 205\div5=41\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 41}\\ {\color{Green} 5}{\overline{\smash{)}205}}\\ -\ 20 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times41=205+0\)

Simplify.

\(\displaystyle 5\times41=205\)

The correct answer is \(\displaystyle 5\times 41\)

Example Question #122 : Grade 6

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 51\\ 5{\overline{\smash{)}255}}\\ -\ 25 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 71\)

\(\displaystyle 5\times 31\)

\(\displaystyle 5\times 52\)

\(\displaystyle 5\times 51\)

\(\displaystyle 5\times 25\)

Correct answer:

\(\displaystyle 5\times 51\)

Explanation:

The computation shows that \(\displaystyle 255\div5=51\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 51}\\ {\color{Green} 5}{\overline{\smash{)}255}}\\ -\ 25 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times51=255+0\)

Simplify.

\(\displaystyle 5\times51=255\)

The correct answer is \(\displaystyle 5\times 51\)

Example Question #3543 : Operations

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 61\\ 5{\overline{\smash{)}305}}\\ -\ 30 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 61\)

\(\displaystyle 5\times 63\)

\(\displaystyle 5\times 31\)

\(\displaystyle 5\times 81\)

\(\displaystyle 5\times 30\)

Correct answer:

\(\displaystyle 5\times 61\)

Explanation:

The computation shows that \(\displaystyle 305\div5=61\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 61}\\ {\color{Green} 5}{\overline{\smash{)}305}}\\ -\ 30 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times61=305+0\)

Simplify.

\(\displaystyle 5\times61=305\)

The correct answer is \(\displaystyle 5\times 61\)

Example Question #3552 : Operations

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 71\\ 5{\overline{\smash{)}355}}\\ -\ 35 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 35\)

\(\displaystyle 5\times 53\)

\(\displaystyle 5\times 51\)

\(\displaystyle 5\times 71\)

\(\displaystyle 5\times 101\)

Correct answer:

\(\displaystyle 5\times 71\)

Explanation:

The computation shows that \(\displaystyle 355\div5=71\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 71}\\ {\color{Green} 5}{\overline{\smash{)}355}}\\ -\ 35 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times71=355+0\)

Simplify.

\(\displaystyle 5\times71=355\)

The correct answer is \(\displaystyle 5\times 71\)

Example Question #31 : Fluently Divide Multi Digit Numbers: Ccss.Math.Content.6.Ns.B.2

Use the computation shown to find the products:

\(\displaystyle \frac{\begin{array}[b]{r} \ 81\\ 5{\overline{\smash{)}405}}\\ -\ 40 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

Possible Answers:

\(\displaystyle 5\times 91\)

\(\displaystyle 5\times 40\)

\(\displaystyle 5\times 101\)

\(\displaystyle 5\times 81\)

\(\displaystyle 5\times 85\)

Correct answer:

\(\displaystyle 5\times 81\)

Explanation:

The computation shows that \(\displaystyle 405\div5=81\) with a remainder of \(\displaystyle 0\).

\(\displaystyle \frac{\begin{array}[b]{r} \ {\color{Green} 81}\\ {\color{Green} 5}{\overline{\smash{)}405}}\\ -\ 40 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }\)

                  \(\displaystyle 0\)

So it must be that:

\(\displaystyle 5\times81=405+0\)

Simplify.

\(\displaystyle 5\times81=405\)

The correct answer is \(\displaystyle 5\times 81\)

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